Numerical methods are the set of tasks by applying arithmetic operations to numerical equations. We can formulate mathematical problems to find the approximate result. This formulation is called the numerical implementation of the problem. In this, there is no need for algorithms. Programming logic is then developed for numerical implementation. The programming is usually done with some high-level languages like Fortran, Basic, etc.
In this article, we will discuss about Bisection Method and Newton Raphson Method as well as the differences between them.
The Bisection Method
The Bisection Method is a numerical technique used to find the root of a continuous function. It is one of the simplest and most reliable methods for solving equations of the form f(x)=0. The method works by repeatedly dividing an interval in half and then selecting the subinterval in which the function changes sign, indicating that a root lies within that subinterval.
Let f(x) is a continuous function in the closed interval [x1, x2], if f(x1), and f(x2) are of opposite signs, then there is at least one root α in the interval (x1, x2), such that f(α) = 0.
Formula for Bisection Method
Formula for Bisection Method is given as:
x_2 = \frac{(x_0 + x_1)} { 2}
Where:
- x0 is the lower bound of the interval.
- x1 is the upper bound of the interval.
- x2 is the midpoint of the interval [x0, x1].
Read more about Bisection method.
Newton Raphson Method
Newton-Raphson Method is a widely used iterative numerical technique for finding the roots of a real-valued function. It is particularly popular because of its fast convergence properties when the initial guess is close to the actual root.
It is the process for the determination of a real root of an equation f(x) = 0 given just one point close to the desired root.
Formula for Newton Raphson Method
Formula for Newton Raphson Method is given as:
x_{n+1} = x_n −\frac { f(x_n)}{f′(x_n)}
Where:
- xn is the current approximation.
- xn+1 is the next approximation.
- f(xn) is the value of the function at xn.
- f′(xn) is the value of the derivative of the function at xn.
For first iteration, we can put n = 0 i.e.,
x_1 = x_0 -\frac{ f(x_0)}{f'(x_0)}
Read more about Newton Raphson Method.
Difference between Bisection Method and Newton Raphson Method
Some of the common key differences between bisection and Newton Raphson method are listed in the following table:
Parameter | Bisection Method | Newton Rapson Method |
|---|---|---|
Rate of Convergence | Linear (slow) | Quadratic (fast) |
Formula used | ||
Initial Approximations | Two initial guesses (x0 and x1) | One initial guess (x0) |
Function Computations per Iteration | 1 | 2 |
Sensitivity to Initial Guess | Less sensitive | Very sensitive |
Derivatives Needed | No | Yes |
Applicability | Not suitable for complex, multiple, or closely spaced roots | Suitable for complex, multiple, or closely spaced roots |
Read More,
- Root Finding Algorithm
- Secant Method
- Trapezoidal Rule
- Simpson's Rule
- LU Decomposition
- Newton Forward And Backward Interpolation
- Lagrange Interpolation
Solved Questions for Bisection Method and Newton Raphson Method
Question 1: Find a root of an equation f(x) = x3 - x - 1 .(Bisection method)
Solution:
Given equation f(x) = x3 - x - 1
let x = 0, 1, 2
In 1st iteration: f(1) = -1 < 0 and f(2) = 5 > 0The root lies between these two points 1 and 2
x0 = 1 + 2/2 = 1.5
f(x0) = f(1.5) = 0.875 > 0In 2nd iteration: f(1) = -1 < 0 and f(1.5) = 0.875 > 0
The root lies between these two points 1 and 1.5
x1 = 1 + 1.5/2 = 1.25
f(x1) = f(1.25) = -0.29688 < 0In 3rd iteration:
f(1.25) = -0.29688 < 0 and f(1.5) = 0.875 > 0The root lies between these two points 1.25 and 1.5
x2 = 1.25 + 1.5/2 = 1.375
f(x2) = f(1.375) = 0.22461 > 0In 4th iteration:
f(1.25) = -0.29688 < 0 and f(1.375) = 0.22461 > 0The root lies between these two points 1.25 and 1.375
x3 = 1.25 + 1.375/2 = 1.3125
f(x3) = f(1.3125) = -0.05151 < 0In 5th iteration:
f(1.3125) = -0.05151 < 0 and f(1.375) = 0.22461 > 0The root lies between these two points 1.3125 and 1.375
x4 = 1.3125 + 1.375/2 = 1.34375
f(x4) = f(1.34375) = 0.08261 > 0In 6th iteration:
f(1.3125) = -0.05151 < 0 and f(1.34375) = 0.08261 > 0The root lies between these two points 1.3125 and 1.34375
x5 = 1.3125 + 1.34375/2 = 1.32812
f(x5) = f(1.32812) = 0.01458 > 0In 7th iteration:
f(1.3125) = -0.05151 < 0 and f(1.32812) = 0.01458 > 0
The root lies between these two points 1.3125 and 1.32812
x6 = 1.3125 + 1.32812/2 = 1.32031
f(x6) = f(1.32031) = -0.01871 < 0In 8th iteration:
f(1.32031) = -0.01871 < 0 and f(1.32812) = 0.01458 > 0The root lies between these two points 1.32031 and 1.32812
x7 = 1.32031 + 1.32812/2 = 1.32422
f(x7) = f(1.32422) = -0.00213 < 0In 9th iteration:
f(1.32422) = -0.00213 < 0 and f(1.32812) = 0.01458 > 0The root lies between these two points 1.32422 and 1.32812
x8 = 1.32422 + 1.32812/2 = 1.32617
f(x8) = f(1.32617) = 0.00621 > 0In 10th iteration:
f(1.32422) = -0.00213 < 0 and f(1.32617) = 0.00621 > 0The root lies between these two points 1.32422 and 1.32617
x9 = 1.32422 + 1.32617/2 = 1.3252
f(x9) = f(1.3252) = 0.00204 > 0In 11th iteration:
f(1.32422) = -0.00213 < 0 and f(1.3252) = 0.00204 > 0The root lies between these two points 1.32422 and 1.3252
x10 = 1.32422 + 1.3252/2 = 1.32471
f(x10) = f(1.32471) = -0.00005 < 0The approximate root of the equation x3 - x - 1 = 0 using the Bisection method is 1.32471
Question 2 : Find a root of an equation f(x) = x3 - x - 1.( Newton Raphson Method)
Solution:
Given equation x3 - x - 1 = 0
Using differentiate method the equation is
∴ f′(x) = 3x2 - 1
Here f(1) = -1 < 0 and f(2) = 5 > 0∴ Root lies between 1 and 2
x0 = 1 + 2/ 2 = 1.5In 1st iteration:
f(x0) = f(1.5) = 0.875
f′(x0) = f′(1.5) = 5.75
x1 = x0 - f(x0) / f′(x0)
x1 = 1.5 - 0.875/ 5.75
x1 = 1.34783In 2nd iteration:
f(x1) = f(1.34783) = 0.10068
f′(x1) = f′(1.34783) = 4.44991
x2 = x1 - f(x1)/f′(x1)
x2 = 1.34783 - 0.10068/4.44991
x2 = 1.3252In 3rd iteration:
f(x2) = f(1.3252) = 0.00206
f′(x2) = f′(1.3252) = 4.26847
x3 = x2 - f(x2)/f′(x2)
x3 = 1.3252 - 0.00206/4.26847
x3 = 1.32472In 4th iteration:
f(x3) = f(1.32472) = 0
f′(x3) = f′(1.32472) = 4.26463
x4 = x3 - f(x3)/f′(x3)
x4 = 1.32472 - 0/ 4.26463
x4 = 1.32472The Approximate root of the equation x3 - x - 1 = 0 using the Newton Raphson method is 1.32472
Practice Problems on Bisection Method and Newton Raphson Method
Problem 1: Use the bisection method to find the root of the function f(x) = x3 − 4x − 9 within the interval [2, 3]. Perform at least 5 iterations.
Problem 2: Apply the bisection method to find a root of f(x) = cos(x)−x in the interval [0, 1]. Carry out the method until the interval width is less than 0.01.
Problem 3: Determine the root of f(x) = ex −3x using the bisection method within the interval [0, 1]. Perform the method until the approximate error is less than 0.001.
Problem 4: Use the Newton-Raphson method to find the root of f(x) = x2 − 2x − 3 starting from an initial guess x0 = 4. Perform at least 4 iterations.
Problem 5: Apply the Newton-Raphson method to find the root of f(x) = x3 − 2x + 1 starting with an initial guess x0 = 0.5. Carry out 5 iterations.
Problem 6: Determine the root of f(x) = tan(x) − x using the Newton-Raphson method, starting from x0 = 4.5. Perform the method until the approximate error is less than 0.0001.